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#### chisigma

##### Well-known member

- Feb 13, 2012

- 1,704

$$\int_{0}^{\infty} \frac{\ln x}{x^{2}+ a^{2}}\ d x\ (1)$$

Scope of this note is to illustrate a general procedure to engage integrals like (1) in elementary way, i.e. without use comnplex analysis tecniques. The preliminary is the evaluation of the following indefinite integral, which doesn't appear in most 'Integration Manuals'...

$$\int x^{m}\ \ln^{n} x\ d x\ (2)$$

... where m and n are non negative integers. Proceeding with standard integration by part we first obtain...

$$\int x^{m}\ \ln^{n} x\ d x = \frac{x^{m+1}}{m+1}\ \ln^{n} x - \frac{n}{m+1}\ \int x^{m}\ \ln^{n-1} x\ d x\ (3)$$

... so that the integral in second term is the original integral with exponent of the logarithm is n-1 instead of n. Repeating n times this procedure we arrive to the final result...

$$\int x^{m}\ \ln^{n} x\ d x = x^{m+1}\ \sum_{i=0}^{n} (-1)^{i}\ \frac{n!}{(n-i)!\ (m+1)^{i+1}}\ \ln^{n-i} x + c\ (4)$$

The (4) is itself important and it will be the basis of all successive steps....

Comments or question about this thread can be submitted in a specific thread open in the 'Commentary Forum'...

http://mathhelpboards.com/commentary-threads-53/commentary-integrals-natural-logarithm-5287.html

Kind regards

$\chi$ $\sigma$